 # Find the limit lim_{urightarrow1}frac{u^4-1}{u^3-1} shadsiei 2020-12-16 Answered
Find the limit
$\underset{u\to 1}{lim}\frac{{u}^{4}-1}{{u}^{3}-1}$
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Given,
$\underset{u\to 1}{lim}\frac{{u}^{4}-1}{{u}^{3}-1}$
On simplification, we get
$\underset{u\to 1}{lim}\frac{{u}^{4}-1}{{u}^{3}-1}=\underset{u\to 1}{lim}\frac{\left({u}^{2}{\right)}^{2}-{1}^{2}}{{u}^{3}-{1}^{3}}$
$=\underset{u\to 1}{lim}\frac{\left({u}^{2}-1\right)\left({u}^{2}+1\right)}{\left(u-1\right)\left({u}^{2}+u+1\right)}$
$\left[\begin{array}{c}\because {a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)\\ {a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)\end{array}\right]$
$=\underset{u\to 1}{lim}\frac{\left(u-1\right)\left(u+1\right)\left({u}^{2}+1\right)}{\left(u-1\right)\left({u}^{2}+u+1\right)}$
$=\underset{u\to 1}{lim}\frac{\left(u+1\right)\left({u}^{2}+1\right)}{\left({u}^{2}+u+1\right)}$
$=\underset{u\to 1}{lim}\frac{\left(1+1\right)\left({1}^{2}+1\right)}{\left({1}^{2}+1+1\right)}$
$=\frac{4}{3}$